Optimaltransportand non-branchinggeodesics · Lp-Monge-Kantorovichproblem Showthattheminimumof inf...

Optimal transport andnon-branching geodesics

Martin Kell

Pisa, November 15th 2018

Lp-Monge problem

• Solve for good µ and arbitrary ν the following

infν=T∗µ

∫dp(x,T (x))dµ(x)

• When is the solution unique?

Optimal transport and non-branching geodesics Martin Kell 1/33

Lp-Monge-Kantorovich problem

• Show that the minimum of

infπ∈Π(µ,ν)

∫dp(x,y)dπ(x,y)

is supported on a graph of measurable map.

• For p= 1 almost never true.• For p ∈ (1,∞) depends on the geometry and on µ.• If true then

• the optimal coupling is unique• Monge = Kantorovich.

infπ∈Π(µ,ν)

∫dp(x,y)dπ(x,y)

infπ∈Π(µ,ν)

∫dp(x,y)dπ(x,y)

infπ∈Π(µ,ν)

∫dp(x,y)dπ(x,y)

Dependence on µ and the geometry ofgeodesics

BLACKBOARD

Non-branching geodesics

• Assumption: (M,d,m) a complete geodesic measure space

Definition (non-branching)A geodesic space (M,d) is non-branching if for all geodesicsγ,η : [0,1]→M with γ0 = η0 and γt = ηt for some t ∈ (0,1) itholds γt = ηt for all t ∈ [0,1].

Equivalently:If m is a midpoint of (x,y) and (x,z) then y = z.

Non-branching and optimal transport

Lemma (no intermediate overlap)A geodesic space is non-branching if the following holds:Whenever for two geodesics γ and η satisfy

dp(γ0,γ1) +dp(η0,η1)≤ dp(γ0,η1) +dp(η0,γ1)

and γt = ηt for some t ∈ (0,1) then γ ≡ η.

Examples of non-branching space

• Riemannian/Finsler manifolds (geodesic = “ODE solution”)• Alexandrov spaces (comparison condition)• Busemann G-spaces (unique continuation property)• CAT(κ)⊕RCD(K,N)-space [Kapovich-Ketterer ’17]

=⇒ works also for MCPloc(K,N)-spaces that are (locally)Busemann convex

• subRiemannian Heisenberg(-type) groups [Ambrosio-Rigot ’04]• subRiemannian Engel group [Ardentov-Sachkov ’11,’15]• Open: Ricci limits, RCD-spaces, Carnot groups

Some history of Lp-Monge-Kantorovich, p > 1

• Theorems using Rademacher Theorem

• in Rn [Brenier ’91, Gangbo-McCann ’96]• Riemannian manifolds [McCann ’01, Gigli ’11]• Finsler manifolds [Villani ’09, Ohta ’09]• Heisenberg groups [Ambrosio-Rigot ’04]• nice subRiemannian manifolds [Figalli-Riffort ’10]• Alexandrov spaces [Bertrand ’08/’15, Schultz-Rajala ’18]

• Anyone missing?

• Theorems using optimal transport and non-branching• non-branching CD(K,N)-spaces [Gigli ’12]• strongly non-branching doubling spaces with interpolation

property [Ambrosio-Rajala ’14]• non-branching spaces with very weak MCP

[Cavalletti-Huesmann ’15]

• using weaker essentially non-branching (e.n.b.) condition• strong CD(K,∞)-spaces [Rajala-Sturm ’14]• RCD(K,N)-spaces [Gigli-Rajala-Sturm ’16]• e.n.b. MCP(K,N)-spaces [Cavalletti-Mondino ’17]• e.n.b. spaces with very weak MCP [K. ’17]

• Theorems using optimal transport and non-branching• non-branching CD(K,N)-spaces [Gigli ’12]• strongly non-branching doubling spaces with interpolation

property [Ambrosio-Rajala ’14]• non-branching spaces with very weak MCP

[Cavalletti-Huesmann ’15]

• using weaker essentially non-branching (e.n.b.) condition• strong CD(K,∞)-spaces [Rajala-Sturm ’14]• RCD(K,N)-spaces [Gigli-Rajala-Sturm ’16]• e.n.b. MCP(K,N)-spaces [Cavalletti-Mondino ’17]• e.n.b. spaces with very weak MCP [K. ’17]

Notation

• Γ⊂M ×M

Γt := {γt |γ ∈Geo[0,1](M,d),(γ0,γ1) ∈ Γ}

• A⊂M and x ∈M

At,x = (A×{x})t

RemarkIn the following fix a p ∈ (1,∞) so that optimal = dp-optimal,cylically monotone = dp-cylically monotone.

Very weak MCP condition

Definition ([Cavalletti-Huesmann ’15])A metric measure space is qualitatively non-degenerate if for allR> 0 there is a function fR : (0,1)→ (0,∞) withCR = limsupt→0 fR(t)> 1

2 such that whenever {x},A⊂BR(x0)then

m(At,x)≥ fR(t)m(A).

RemarkNote that 2CR > 1.

Optimal transport maps

Definition (Good Transport Behavior)A metric measure space (M,d,m) has good transport behavior(GTB)p if for all µ ∈ Pacp (M) and all ν ∈ Pp(M) every optimalcoupling π is induced by a transport map T , i.e. π = (id×T )∗µ.

Theorem ([Cavalletti-Huesmann ’15])Assume (M,d,m) is qualitatively non-degenerate andnon-branching. Then (M,d,m) has good transport behavior(GTB)p for all p ∈ (1,∞).

Proof for ν = (1−λ)δx1 +λδx2, x1 6= x2

• Choose some optimal coupling π and note

suppπ =A1×{x1}∪̇A2×{x2}∪̇A×{x1,x2}.

• Observation:• π is induced by a transport map iff m(A) = 0.• by non-branching for t ∈ (0,1)

At,x1 ∩At,x2 = ∅.

• by qualitative non-degeneracy (and A is compact)

m(A)≥ limsupt→0

m(At,x1 ∪At,x2)

= limsupt→0

m(At,x1) +m(At,x2)

= 2limsupt→0

f(t)m(A) = 2CRm(A).

• Conclusion: m(A) = 0 and π is induced by a transport map.Optimal transport and non-branching geodesics Martin Kell 12/33

suppπ =A1×{x1}∪̇A2×{x2}∪̇A×{x1,x2}.

At,x1 ∩At,x2 = ∅.

m(A)≥ limsupt→0

m(At,x1 ∪At,x2)

= limsupt→0

m(At,x1) +m(At,x2)

= 2limsupt→0

f(t)m(A) = 2CRm(A).

suppπ =A1×{x1}∪̇A2×{x2}∪̇A×{x1,x2}.

At,x1 ∩At,x2 = ∅.

m(A)≥ limsupt→0

m(At,x1 ∪At,x2)

= limsupt→0

m(At,x1) +m(At,x2)

= 2limsupt→0

f(t)m(A) = 2CRm(A).

suppπ =A1×{x1}∪̇A2×{x2}∪̇A×{x1,x2}.

At,x1 ∩At,x2 = ∅.

m(A)≥ limsupt→0

m(At,x1 ∪At,x2)

= limsupt→0

m(At,x1) +m(At,x2)

= 2limsupt→0

f(t)m(A) = 2CRm(A).

suppπ =A1×{x1}∪̇A2×{x2}∪̇A×{x1,x2}.

At,x1 ∩At,x2 = ∅.

m(A)≥ limsupt→0

m(At,x1 ∪At,x2)

= limsupt→0

m(At,x1) +m(At,x2)

= 2limsupt→0

f(t)m(A) = 2CRm(A).

suppπ =A1×{x1}∪̇A2×{x2}∪̇A×{x1,x2}.

At,x1 ∩At,x2 = ∅.

m(A)≥ limsupt→0

m(At,x1 ∪At,x2)

= limsupt→0

m(At,x1) +m(At,x2)

= 2limsupt→0

f(t)m(A) = 2CRm(A).

suppπ =A1×{x1}∪̇A2×{x2}∪̇A×{x1,x2}.

At,x1 ∩At,x2 = ∅.

m(A)≥ limsupt→0

m(At,x1 ∪At,x2)

= limsupt→0

m(At,x1) +m(At,x2)

= 2limsupt→0

f(t)m(A) = 2CRm(A).

Proof for ν = ∑ni=1λiδxi

• Choose some optimal coupling π then

suppπ =n⋃i=1

Ai×{xi}∪̇n⋃i<j

Aij×{xi,xj}.

• By previous slide m(Aij) = 0.• Hence

T (x) ={xi x ∈Aix otherwise

is a transport map.

suppπ =n⋃i=1

Aij×{xi,xj}.

is a transport map.

suppπ =n⋃i=1

Aij×{xi,xj}.

is a transport map.

suppπ =n⋃i=1

Aij×{xi,xj}.

is a transport map.

Observation

• For distinct x,y ∈M and compact A⊂M

A= = {z ∈M |d(x,z) = d(y,z)}A 6= =A\A=.

• If A×{x,y} is cyclically monotone then even withoutnon-branching

(A 6=)t,x∩ (A 6=)t,y = ∅.

• Hence if

∀x 6= y : m({z ∈M |d(x,z) = d(y,z)}= 0

then any optimal coupling between µ�m and ν discrete isinduced by a transport map.

• This holds for any normed space and measure assigning zeromass to hyperplanes.

Observation

A= = {z ∈M |d(x,z) = d(y,z)}A6= =A\A=.

(A 6=)t,x∩ (A 6=)t,y = ∅.

• Hence if

∀x 6= y : m({z ∈M |d(x,z) = d(y,z)}= 0

Observation

A= = {z ∈M |d(x,z) = d(y,z)}A6= =A\A=.

(A 6=)t,x∩ (A 6=)t,y = ∅.

• Hence if

∀x 6= y : m({z ∈M |d(x,z) = d(y,z)}= 0

Observation

A= = {z ∈M |d(x,z) = d(y,z)}A6= =A\A=.

(A 6=)t,x∩ (A 6=)t,y = ∅.

• Hence if

∀x 6= y : m({z ∈M |d(x,z) = d(y,z)}= 0

Observation

A= = {z ∈M |d(x,z) = d(y,z)}A6= =A\A=.

(A 6=)t,x∩ (A 6=)t,y = ∅.

• Hence if

∀x 6= y : m({z ∈M |d(x,z) = d(y,z)}= 0

Proof for general ν

Theorem (Selection dichotomy, see e.g. [K. ’17])If some optimal coupling π is not induced by a transport map thenthere is a compact set K ⊂ supp(p1)∗π with π(K×M)> 0 andtwo continuous maps T1,T2 :M →M with T1(K)∩T2(K) = ∅such that

Γ(1)∪Γ(2)

is cyclically monotone where Γ(i) = graphK Ti.

LemmaIf (M,d) is non-branching then

Γ(1)t ∩Γ(2)

t = ∅

andm(K)≥ limsup

[m(Γ(1)

t ) +m(Γ(2)t )]

Γ(1)∪Γ(2)

Γ(1)t ∩Γ(2)

t = ∅

andm(K)≥ limsup

[m(Γ(1)

t ) +m(Γ(2)t )]

Γ(1)∪Γ(2)

Γ(1)t ∩Γ(2)

t = ∅

andm(K)≥ limsup

[m(Γ(1)

t ) +m(Γ(2)t )]

Uniform approximation (simplified for presentation)

LemmaIf (M,d,m) is non-branching and qualitatively non-degenerate then

m(Γ(i)t )≥ f(t)m(K).

Idea of proof.Let νn→ (Ti)∗µ

∣∣K

with νn discrete then eventually

Γ(i),nt

!⊂ (Γ(i)

so thatm(Γ(i)

t )≥ limsupn→0

m(Γ(i),nt )≥ f(t)m(K).

Uniform approximation (simplified for presentation)

LemmaIf (M,d,m) is non-branching and qualitatively non-degenerate then

m(Γ(i)t )≥ f(t)m(K).

Idea of proof.Let νn→ (Ti)∗µ

∣∣K

with νn discrete then eventually

Γ(i),nt

!⊂ (Γ(i)

so thatm(Γ(i)

t )≥ limsupn→0

m(Γ(i),nt )≥ f(t)m(K).

Conclusion

• If the claim was wrong then using we arrive at the followingcontradiction

m(K)≥ limsupt→0

m(Γ(1)t ) +m(Γ(2)

≥ 2limsupt→0

f(t)m(K) = 2CRm(K).

• Thus, any optimal transport between µ�m and arbitrary νis induced by a transport map.

Conclusion

• If the claim was wrong then using we arrive at the followingcontradiction

m(K)≥ limsupt→0

m(Γ(1)t ) +m(Γ(2)

≥ 2limsupt→0

f(t)m(K) = 2CRm(K).

• Thus, any optimal transport between µ�m and arbitrary νis induced by a transport map.

Ingredients of the proof

• (weak) non-branching property

m(Γ(1)t ∩Γ(2)

t ) = 0

m(Γ(1)t ∩Γ(2)

t ) = 0

m(Γ(1)t ∩Γ(2)

t ) = 0

• qualitative non-degeneracy

m(Γ(1)t ∩Γ(2)

t ) = 0

• qualitative non-degeneracy =⇒ not an optimal transportproperty

Partial extension to p= 1, see e.g. [K.-Suhr 18]

• For simplicity let M = Rn

LemmaIf Γ⊂ {xn < 0}×{xn = 0} is d-cyclically monotone then for alldistinct geodesics γ,η with

(γ0,γ1),(η0,η1) ∈ Γ

it holds γt 6= ηt.

CorollaryAssume suppµ× suppν ⊂ {xn < 0}×{xn = 0} then anyd-optimal coupling is induced by a transport map.

RemarkWorks for non-bran., qual. non-deg. spaces if ν is supported in alevel set of a dual solution.

(γ0,γ1),(η0,η1) ∈ Γ

Lorentzian setting [K.-Suhr ’18]

• Lagrangian for p ∈ (0,1]

Lp(v) ={−(−g(v,v))

p2 v ∈ C̄

∞ otherwise

induces cost function cp :M ×M → (−∞,0]∪{∞}• For p ∈ (0,1), geodesics connecting (x,y) ∈ c−1

p ((−∞,0)) arenon-branching.

• For p= 1 and hyperbolic spacetimes, introduce smooth timefunction τ :M → R with

∀v ∈ C̄ : dτ(v)> 0

and then all causal geodesics can be parametrizedtime-affinely and geodesics with endpoints in

{τ < a}×{τ = a}

are non-branching.Optimal transport and non-branching geodesics Martin Kell 20/33

Lp(v) ={−(−g(v,v))

p2 v ∈ C̄

∞ otherwise

∀v ∈ C̄ : dτ(v)> 0

{τ < a}×{τ = a}

Lp(v) ={−(−g(v,v))

p2 v ∈ C̄

∞ otherwise

∀v ∈ C̄ : dτ(v)> 0

{τ < a}×{τ = a}

Spaces with good transport behavior (I)

• Assume (M,d,m) has good transport behavior (GTB)p

• for every µ0�m and µ1 the optimal coupling is unique andinduced by a transport map

• for every µ0�m and µ1 the geodesic t 7→ µt is unique

• if, in addition, m is qual. non-deg. then µt�m

Obstructions to good transport behavior (I)

• Let T = [0,1]1∪ [0,1]3∪ [0,1]3 be the tripod glued at 0.

• (T,d,m) is a CAT (0)-space, but will never have goodtransport behavior.

Obstructions to good transport behavior (I)

• Let T = [0,1]1∪ [0,1]3∪ [0,1]3 be the tripod glued at 0.

• (T,d,m) is a CAT (0)-space, but will never have goodtransport behavior.

Essentially non-branching (I)

• A subset of geodesics G ⊂Geo[0,1](M,d) is callednon-branching if for all γ,η ∈ G with γ0 = η0 or γ1 = η1 itholds whenever γt = ηt for some t ∈ (0,1) it holds γt = ηt forall t ∈ [0,1].

Definition ([Rajala-Sturm ’14])The space (M,d,m) is essentially non-branching (ENB)p if forevery optimal dynamical coupling σ with (e0)∗σ,(e1)∗σ�m isconcentrated on a non-branching set.

• May alter condition: For each t1, . . . , tn ∈ (0,1), σ isconcentrated on a cyclically monotone set Γ such that for alldistinct geodesics γ and η with endpoints in Γ it holdsγti 6= ηti .

Spaces with good transport behavior (II)

• A subset of geodesics G ⊂Geo[0,1](M,d) is callednon-branching to the left if for all γ,η ∈ G with γ0 = η0 itholds whenever γt = ηt for some t ∈ (0,1) it holds γt = ηt forall t ∈ [0,1].

Theorem ([K. ’17])If (M,d,m) has good transport behavior then any dynamicalcoupling σ with (e0)∗σ�m is concentrated on a set that isnon-branching to the left. In particular, (M,d,m) is essentiallynon-branching (ENB)p.

Spaces with good transport behavior (II)

• A subset of geodesics G ⊂Geo[0,1](M,d) is callednon-branching to the left if for all γ,η ∈ G with γ0 = η0 itholds whenever γt = ηt for some t ∈ (0,1) it holds γt = ηt forall t ∈ [0,1].

Theorem ([K. ’17])If (M,d,m) has good transport behavior then any dynamicalcoupling σ with (e0)∗σ�m is concentrated on a set that isnon-branching to the left. In particular, (M,d,m) is essentiallynon-branching (ENB)p.

Obstructions to good transport behavior (II)

• Choose three mutually singular measures m1,m2 and m3 on[0,1].

• For the tripod (T,d), regard them as measures on [0,1i] andset m = m1 +m2 +m3

• Observations:• (T,d,m) is essentially non-branching• all optimal couplings π with (pi)∗π�m are induced by

transport map

Obstructions to good transport behavior (II)

• Choose three mutually singular measures m1,m2 and m3 on[0,1].

• For the tripod (T,d), regard them as measures on [0,1i] andset m = m1 +m2 +m3

• Observations:• (T,d,m) is essentially non-branching• all optimal couplings π with (pi)∗π�m are induced by

transport map

Essentially non-branching (II)

Theorem ([K. ’17])If (M,d,m) is essentially non-branching (ENB)p and qualitativelynon-degenerate then it has good transport behavior (GTB)p.

Idea of the proof

• As ν 6�m essentially non-branching cannot be directly used• Construct dynamical optimal coupling σ with

(eε)∗σ,(e1−ε)∗σ�m with

m∣∣Γ ε

1−ε� (eε)∗σ�m

∣∣Γ ε

1−ε

for a cyclically monotone set Γ with (e0,e1−ε)∗σ(Γ) = 1.• Essentially non-branching implies

m(Γ(1)ε ∩Γ(2)

ε ) = 0

when Γ(1)is given via the Selection Dichotomy.• A proof a la Cavalletti–Huesmann gives the result.

Idea of the proof

m∣∣Γ ε

1−ε� (eε)∗σ�m

∣∣Γ ε

1−ε

m(Γ(1)ε ∩Γ(2)

ε ) = 0

Idea of the proof

m∣∣Γ ε

1−ε� (eε)∗σ�m

∣∣Γ ε

1−ε

m(Γ(1)ε ∩Γ(2)

ε ) = 0

Idea of the proof

m∣∣Γ ε

1−ε� (eε)∗σ�m

∣∣Γ ε

1−ε

m(Γ(1)ε ∩Γ(2)

ε ) = 0

Idea of the proof

m∣∣Γ ε

1−ε� (eε)∗σ�m

∣∣Γ ε

1−ε

m(Γ(1)ε ∩Γ(2)

ε ) = 0

Intermediate summary

• Assume (M,d,m) essentially non-branching and qualitativelynon-degenerate

• Let µ0 = f0m and µ1 arbitrary• Conclusion:

1 (uniqueness) unique optimal dynamical coupling σ2 (good transport behavior)

(e0,e1)∗σ = (id×T1)∗µ0

3 (strong interpolation property)

(et)∗σ = ftm

4 (strong bounded density property)

ft(γt)≤1

fR(t)f0(γ0).

Application to Measure Rigidity

Theorem ([K. ’17])Assume (M,d,m1) and (M,d,m2) are both essentiallynon-branching (ENB)p and qualitatively non-degenerate then m1and m2 are mutually absolutely continuous.

CorollaryFor i= 1,2 let (M,d,mi) be RCD(Ki,Ni)-spaces with Ni <∞.Then m1 and m2 are mutually absolutely continuous.

1st Proof of the Measure Rigidity (I)

• Decompose m2 = fm1 +ms2

• Assume, by contradiction, ms2 6= 0.

• Observation: We must have f 6= 0 by strong interpolationproperty

• The following claim implies gives a contradiction

Claimms

2 is both essentially non-branching and qualitativelynon-degenerate

Claimms

1st Proof (II) - proof of the claim

• Note if m1(At,x) = m1(A) = 0 and then

ms2(At,x) = m2(At,x)≥ fR(t)m2(A) = fR(t)ms

• Observation: Since m2(At,x)> 0 for all t ∈ (0,1], it is possibleto show that for m2-a.e. x ∈A there is a unique geodesicγ(x) such that

x ∈ γ(x)((0,1))

LemmaThere is K ⊂⊂A with m2(K)> 0 and t 7→ µt geodesic withµ1�m2, µt0 = 1

m2(K)m2∣∣K

and µ1 = δx.

• Lemma implies if µt0⊥m1 then µt⊥m1 for all t ∈ [0,1) whichyields the claim.

x ∈ γ(x)((0,1))

m2(K)m2∣∣K

and µ1 = δx.

x ∈ γ(x)((0,1))

m2(K)m2∣∣K

and µ1 = δx.

x ∈ γ(x)((0,1))

m2(K)m2∣∣K

and µ1 = δx.

2nd Proof of the Measure Rigidity

• Assume mss 6= 0 and choose µ0 = 1

ms2(K)ms

2∣∣K

and µ1�m1

• By strong intersection property

µt�m1,µt�m2

hence µt⊥ms2

• By bounded density property for µt = ftm2

‖ft‖∞ ≤1

fR(t)ms2(K)

• Arrive at contradiction using the following lemma.

Lemma (Self-intersection [CH ’14, K. ’17])If µ= 1

m(K)m∣∣K

and µn = fnm with Wp(µn,µ)→ 0 and‖fn‖∞ ≤ C then µ 6 ⊥µn for all sufficiently large n.

ms2(K)ms

2∣∣K

and µ1�m1

µt�m1,µt�m2

hence µt⊥ms2

‖ft‖∞ ≤1

fR(t)ms2(K)

m(K)m∣∣K

ms2(K)ms

2∣∣K

and µ1�m1

µt�m1,µt�m2

hence µt⊥ms2

‖ft‖∞ ≤1

fR(t)ms2(K)

m(K)m∣∣K

ms2(K)ms

2∣∣K

and µ1�m1

µt�m1,µt�m2

hence µt⊥ms2

‖ft‖∞ ≤1

fR(t)ms2(K)

m(K)m∣∣K

ms2(K)ms

2∣∣K

and µ1�m1

µt�m1,µt�m2

hence µt⊥ms2

‖ft‖∞ ≤1

fR(t)ms2(K)

m(K)m∣∣K

Thank you for your attention

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